2025高考数学一轮复习-4.2-同角三角函数的基本关系与诱导公式-专项训练【含答案】

2025高考数学一轮复习-4.2-同角三角函数的基本关系与诱导公式-专项训练【A级基础巩固】1.已知sin(α-π4)=13,则cos(π4A.223 B.-223 C.2.已知3sin(θ+π2)+sin(θ+π)=0,θ∈(-π,0),则sinθ等(A.-31010 B.-1010 C.33.已知α∈(0,π),且2cos2α-sin2α-4cosα=3,则sinα等于()A.53 B.23 C.134.(多选题)在△ABC中,下列结论正确的是()A.sin(A+B)=sinCB.sinB+CC.tan(A+B)=-tanC(C≠π2D.cos(A+B)=cosC5.已知角α的顶点在原点,始边与x轴非负半轴重合,终边与直线2x+y+3=0平行,则sinαA.-2 B.-14 C.2 6.已知tanθ=2,则1+sinθcosθ的值为.

7.已知角θ的终边过点A(4,a),且sin(θ-π)=35,则tanθ等于8.若sinx=3sin(x-π2),则cosxcos(x+π2)=9.已知角θ的终边与单位圆x2+y2=1在第四象限交于点P,且点P的坐标为(12(1)求tanθ的值;(2)求cos(πINCLUDEPICTURE"B组.TIF"INCLUDEPICTURE"E:\\大样\\人教数学\\B组.TIF"INET【B级能力提升】10.已知tanα=-3,则sinA.-34 B.34 C.31011.(多选题)已知3sin(π+θ)=cos(2π-θ),θ∈(-π3A.-π6 B.-π3 C.π312.已知sin(3π+θ)=13,则cos(π+.

13.已知sin(π-α)-cos(π+α)=23(1)sinα-cosα;(2)sin3(π+α)+cos3(-α).INCLUDEPICTURE"B组.TIF"INCLUDEPICTURE"E:\\大样\\人教数学\\B组.TIF"INET【C级应用创新练】14.已知f(α)=1+sinα1-(1)若f(α)=3,求43sin2α+cos2(2)若cos2αf(α)=12,求cos(2023π+α)+cos(3π参考答案【A级基础巩固】1.解析:因为π4+α-(α-π4)=所以cos(π4+α)=cos[π2+(α-π4)]=-sin(α-π2.解析:因为3sin(θ+π2)+sin(θ+π)=0,所以3cosθ-sinθ=0,因为θ∈(-π,0),sin2θ+cos2θ=1,所以sinθ=-33.解析:因为2cos2α-sin2α-4cosα=3,所以3cos2α-4cosα-4=0,解得cosα=-231-cos2α4.解析:在△ABC中,有A+B+C=π,则sin(A+B)=sin(π-C)=sinC,A正确;sinB+C2=sin(π2-(C≠π25.解析:因为角α的终边与直线2x+y+3=0平行,即角α的终边在直线y=-2x上,所以tanα=-2,所以sinα-cos6.解析:因为tanθ=2,所以1+sinθcosθ=sintan2θ+tanθ答案:77.解析:sin(θ-π)=-sinθ=35,sinθ=-3由于角θ的终边过点A(4,a),所以θ在第四象限,所以cosθ=1-sin所以tanθ=sinθcosθ答案:-38.解析:因为sinx=3sin(x-π2)=-3cosx,所以sin2x=9cos2x=1-cos2所以cos2x=110所以cosxcos(x+π2)=-cosxsinx=(-cosx)·=3cos2x=3×110=3答案:39.解:(1)由θ为第四象限角,终边与单位圆交于点P(12得(12)2+y2解得y=-32所以tanθ=-321(2)因为tanθ=-3,所以cos(π2-θ)+cos(θ-2πINCLUDEPICTURE"B组.TIF"INCLUDEPICTURE"E:\\大样\\人教数学\\B组.TIF"INET【B级能力提升】10.解析:因为tanα=-3,所以sin3α-sin-sinαcosα=-sinαcosαco11.解析:因为3sin(π+θ)=cos(2π-θ),所以-3·sinθ=cosθ,所以tanθ=-33,因为θ∈(-π3,π),所以θ=-π612.解析:由sin(3π+θ)=13,可得sinθ=-1所以cos(π+=-cosθ=11+cosθ=2=21-=18.答案:1813.解:(1)由sin(π-α)-cos(π+α)=23,得sinα+cosα=2所以1+2sinαcosα=29,故2sinαcosα=-7所以sinα>0,cosα<0,所以sinα-cosα>0,所以sinα-cosα=(sinα-cosα)2(2)sin3(π+α)+cos3(-α)=cos3α-sin3α=(cosα-sinα)(cos2α+cosαsinα+sin2α)=(-43)×(1-718)=-INCLUDEPICTURE"B组.TIF"INCLUDEPICTURE"E:\\大样\\人教数学\\B组.TIF"INET【C级应用创新练】14.解:(1)因为α为第二象限角,所以|cosα|=-cosα,f(α)=1+sinα1=(1+sinα=(1+sinα=1+sinα|cosα|若f(α)=3,则-2tanα=3,所以tanα=-32所以43sin2α+cos2α=43sin2α+co(2)cos2αf(α)=cos2α×(-2tanα)=-cos2α×2sinα因为